Fundamentals of Electrical Drivess - Chapter 4 doc

Chapter 4 THREE-PHASE CIRCUITS 4.1 Introduction The majority of electrical drive systems in use are powered by a so-called three-phase (three wire) supply. The main reason for this is that a more efficient energy transfer from supply to the load, such as a three-phase AC machine, is possible in comparison with a single (two wire) AC circuit. The load, being the machine acting as a motor, is formed by three phases. Each phase-winding of which has two terminals, yielding a total of six terminal-bolts, usually config- ured as sketched in figure 4.1. The phase impedances are assumed to be equal. Figure 4.1. Connector on a three phase machine, with a to- tal of six terminals The terminal lay-out as shown in figure 4.1 has been purposely chosen to al- low the user to readily connect the machine’s phase windings in two distinct configurations. The star and delta configurations are depicted in figure 4.3 and figure 4.9 respectively. Voltages and currents in the different configurations are identified by the subscripts S1, S2, S3 when the machine is in star, wye or Y-configuration. Subscripts D1, D2, D3 apply to the delta or ∆ configuration. The voltages/currents, identified by the subscripts R, S, T are linked to the supply source, which is usually a power electronic converter or the three-phase grid. Figure 4.2 shows an example of a three-phase voltage supply which gener- ates three voltages (of arbitrary shape) u R , u S , u T that are defined with respect to the 0V (neutral) of this system. In this chapter we will look into modelling 76 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 4.2. Supply conven- tion (voltage sources shown) three-phase circuits, and in this context introduce a new set of building blocks as required to move (in both directions) from machine phase variables to supply variables for either star or delta connected machines. So-called space vectors are introduced as an important tool to simplify the dynamic analysis of three-phase circuits. In the sequel to this chapter the link between phasors and space vectors is made in order to examine three- phase circuits under steady-state conditions in case the supply is deemed to be sinusoidal in nature. Finally, a set of tutorials will be provided which serves to reinforce the concepts outlined in this chapter. 4.2 Star/Wye connected circuit The term ‘star’ or ‘wye’ connected circuit refers to the configuration shown in figure 4.4, where the machine phases are connected in such a manner that a common ‘star’ or ‘neutral’ point is established. This star-point is usually not connected to the neutral or 0V reference point of the supply. For the ‘star’ connected configuration the lower three terminals v 2 ,w 2 ,u 2 are interconnected as shown by the red lines in figure 4.3. This figure also shows how the R, S and T supply is connected to the machine terminals. The supply voltages u R , u S , u T and u S0 are defined with respect to the 0V of the supply source (see figure 4.2). Note (again) that the supply voltages are instantaneous functions of time and need not be sinusoidal. Furthermore, the sum of the three voltages does not and indeed will not usually be zero when a power electronic converter is used as a supply source. On the basis of Kirchhoff’s voltage and current laws and observation of figure 4.4 we will determine the relationships that exist between supply and phase variables. With respect to the phase variables the following expressions are valid i S1 + i S2 + i S3 =0 (4.1a) u S1 + u S2 + u S3 =0 (4.1b) Three-phase Circuits 77 Figure 4.3. Three phase ma- chine, star (Y) connected Figure 4.4. Star/Wye con- nected according figure 4.3 Note that equation (4.1b) shows that the sum of the phase voltages is zero. This is indeed the case here because the phase impedances are deemed to be equal. The supply currents i R , i S , i T are in this case equal to the phase currents i S1 , i S2 , i S3 respectively. Hence, the building block as shown in figure 4.5(a) has a transfer function as given by equation (4.2).   i S1 i S2 i S3   =   100 010 001     i R i S i T   (4.2) In the following analysis we will also directly discuss the inverse, i.e. the transfer function and building block(s) needed to return from phase to supply variables. This approach is instructive as cascading the two modules must give the original supply waveforms. In this case the inverse is the unity matrix as 78 FUNDAMENTALS OF ELECTRICAL DRIVES (a) Supply to phase (b) Phase to supply Figure 4.5. Current conversions: star connected represented by equation (4.3) andbuilding blockas represented by figure 4.5(b).   i R i S i T   =   100 010 001     i S1 i S2 i S3   (4.3) The conversion of supply to phase voltages is according to figure 4.4 of the form given by,   u S1 u S2 u S3   =   100 010 001     u R u S u T   −   1 1 1   u S0 (4.4) in which the voltage u 0 given in equation (4.4) is the potential of the star point with respect to the 0V reference of the supply. The voltage u S0 is the so-called zero sequence component and can be found with the aid of equations (4.1b), (4.4) which leads to u S0 = u R + u S + u T 3 (4.5) The conversion module whichrepresents equation(4.4) is given byfigure 4.6(a). An important observation from figure 4.6(a) is that this module has a fourth out- put, the voltage u S0 , which is obtained from u R , u S , u T and the superposition (a) Supply to phase (b) Phase to supply Figure 4.6. Voltage conversions: star connected Three-phase Circuits 79 of symmetrical machine impedances. The inversion follows directly from fig- ure 4.4 and is of the form   u R u S u T   =   100 010 001     u S1 u S2 u S3   +   1 1 1   u S0 (4.6) In equation(4.6) the valueof u S0 can bechosen freely, hence thesupply voltages u R , u S , u T are not unique for a given set of phase voltages u S1 , u S2 , u S3 .The conversion module is given in figure 4.6(b). 4.2.1 Modelling star connected circuit The single phase R, L circuit model has been discussed earlier and the generic implementation given in figure 2.5 on page 32 needs to be duplicated three times, as shown in figure 4.7. Note that the three-phase R-L model shown Figure 4.7. Generic three-phase R-L model in figure 4.7 is a simplified representation of an AC machine. In reality, mutual coupling terms exist between the phases which severely complicates the three- phase circuit model. At a later stage in this chapter an alternative approach to modelling three-phase circuits will be given, which is able to handle more com- plex circuits than the R-L concept considered here. The combined conversion process with all the building blocks needed to arrive at the supply currents, on the basis of a given set of supply voltages, is given in figure 4.8. 80 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 4.8. Star connected circuit model 4.3 Delta connected circuit The term ‘delta’ connected circuit refers to the configuration shown in fig- ure 4.10. In the terminal box on the machine, the terminals pairs (u 1 , v 2 ), (v 1 , w 2 ) and (w 1 , u 2 ) are interconnected, as shown by three red lines in figure 4.9. The delta connection is often used in applications with relatively low supply voltages. Furthermore, delta connected machines are commonly used in high power applications (typically from about 0.5 MW upwards). Figure 4.9. Three phase ma- chine, delta (∆) connected Figure4.10. Deltaconnected according to figure 4.9 The supply voltages u R , u S , u T are defined with respect to the 0V of the supply source inthesamemanner as discussedinsection4.2. It isre-emphasized Three-phase Circuits 81 that the supply voltages are instantaneous functions of time and need not be sinusoidal and the sum of the three do not need to be zero. On the basis of Kirchhoff’s voltage and current laws and observation of figure 4.10 we can again determine the relationships that exist between supply and phase variables. With respect to the phase variables the following expressions are valid i D1 + i D2 + i D3 =3i D0 (4.7a) u D1 + u D2 + u D3 =0 (4.7b) where i D0 represents a so-called zero sequence current. In the circuit model as given infigure 4.10 no such currentwill exist. However, if for example a voltage source is introduced in each phase leg, which has a third harmonic component then anon-zero loop current i D0 will begenerated, hence i D1 = i D0 , i D2 = i D0 and i D3 = i D0 . Under these conditions the sum of these phase currents is equal to 3i D0 as shown by equation (4.7a). Measurements from a practical system with substantial loop-current are shown on page 98. Equation (4.7a) may also be written as i D0 = i D1 + i D2 + i D3 3 (4.8) The relationship between supply currents i R , i S , i T and phase currents i D1 , i D2 , i D3 is in this case found using Kirchhoff’s current law and observation of figure 4.10. For example the current i R may be expressed as i R = i D1 − i D2 . If we extend this analysis to all three phases the transfer function according to equation (4.9) appears.   i R i S i T   =   1 −10 01−1 −101     i D1 i D2 i D3   (4.9) The conversion module which represents equations (4.9), (4.5) is given by figure 4.11(a). (a) Phase to Supply (b) Supply to phase Figure 4.11. Current conversions: delta connected An important observation from figure 4.11(a) is that this module has a fourth output the currenti D0 , as defined by equation (4.8), which isrequired in order to 82 FUNDAMENTALS OF ELECTRICAL DRIVES facilitate the conversion fromphase currents tosupply currents. This conversion follows from figure 4.10 and it is instructive to initially consider the process by which an expression for the branch current i D1 is formed. From figure 4.10 the following expressions can be found i D1 = i R + i D2 (4.10a) i D1 = −i T + i D3 (4.10b) Adding equations (4.10a), (4.10b) gives 2i D1 = i R − i T +(i D2 + i D3 )    −i D1 +3i D0 (4.11) where the term (i D2 + i D3 ) can according to equation (4.7a) also be written as (−i D1 +3i D0 ), which leads to i D1 = 1 3 (i R − i T )+i D0 . It is noted that this expression is in fact not an explicit function for i D1 given that i D0 is also a function of the currents i D1 , i D2 , i D3 . This means that the conversion from supply to phase current can only be made if the current i D0 is known, i.e. obtained from the ‘delta’ phase to supply current conversion module discussed earlier (see figure 4.11(a)). The exception to this rule is the case where the sum of the phase currents will be zero, as is the case when the latter are sinusoidal, of equal amplitude and displaced by an angle of 2π/3 with respect to each other. If we extend this single phase analysis for i D1 to all three phases, the conversion matrix, as given by expression (4.12) and building module (figure 4.11(b)), appears.      i D1 i D2 i D3      =      1 3 0 − 1 3 − 1 3 1 3 0 0 − 1 3 1 3           i R i S i T      +      1 1 1      i D0 (4.12) The conversion of supply voltages tophase voltage isaccording to figure 4.10 of the form given by equation (4.13).   u D1 u D2 u D3   =   10−1 −110 0 −11     u R u S u T   (4.13) The conversion module which represents equation (4.13) is given by fig- ure 4.12(a). Figure 4.12(a) has a fourth output, the voltage u S0 , as found using equation (4.5), which is again required to facilitate the conversion from phase voltageto supplyvoltages. The inversion follows directly from figure (4.10) and can be made more translucent by initially considering a single phase conversion Three-phase Circuits 83 (a) Supply to phase (b) Phase to supply Figure 4.12. Voltage conversions: delta connected first. An observation of figure 4.10 learns that the following two expressions may be found which contain the voltage u R . u R = u D1 + u T (4.14a) u R = −u D2 + u S (4.14b) Adding equations (4.14a), (4.14b) gives 2u R = u D1 − u D2 +(u T + u S )    −u R +3u S0 (4.15) where the term (u T + u S ) can according to equation (4.5) also be written as (−u R +3u S0 ), which leads to u R = 1 3 (u D1 − u D2 )+u S0 . It is noted (as was the case for the ‘current’ conversion) that this expression is in fact not an explicit expression for u R given that u S0 is also a function of the voltages u R , u S , u T . This means that the conversion from phase to supply voltages can only be made if the voltage u S0 is known. In the case where the sum of the supply voltages is zero, as is the case when the latter are sinusoidal, of equal magnitude and displaced by an angle of 2π/3 with respect to each other the voltage u S0 will be zero. If we extend our single phase analysis shown above for u R to the remaining two phases the conversion matrix as given by expression (4.16) and building module (figure 4.12(b)), appears.      u R u S u T      =      1 3 − 1 3 0 0 1 3 − 1 3 − 1 3 0 1 3           u D1 u D2 u D3      +      1 1 1      u S0 (4.16) 4.3.1 Modelling delta connected circuit The three-phase R, L generic circuit model, as shown in figure 4.7, for the star connected phase configuration is directly applicable herewith the important difference that the current/voltage phase variables u S1 , u S2 , u S3 , i S1 , i S2 , i S3 must be replaced by the variables u D1 , u D2 , u D3 , i D1 , i D2 , i D3 given that we are dealing with a delta connected load. The inputs to this module will be the 84 FUNDAMENTALS OF ELECTRICAL DRIVES phase voltages from the delta connected circuit and the outputs are the three phase currents. The conversion process needed to arrive at the supply currents given a set of supply voltages is shown in figure 4.13. Figure 4.13. Delta connected circuit model 4.4 Space vectors The question as to why we need ‘space vectors’ comes down to the difficulty of handling complex three phase systems as was mentioned earlier. It will be shown that the introduction of a space vector type representation for a three- phase system leads to considerable simplification. The space vector formulation is in its general form given by equation (4.17). x = C  x R + x S e jγ + x T e j2γ  (4.17) with γ = 2π 3 . The variables x R , x S , x T represent three instantaneous time de- pendent supply variables. These may for example, be the three supply voltages u R , u S , u T , or in fact any other variable. Furthermore, these variables are real quantities and do not need to be sinusoidal. The constant C is a scalar and its value will be defined at a later stage. The space vector x itself is both complex and time dependent. The space vector is represented in a complex plane which at present is assumed to be stationary. The space vector can according to equation (4.18) also be written in terms of a real x α and imaginary x β component with j = √ −1. x = x α + jx β (4.18) Figure 4.14 shows the space vector in the complex plane. Note that x α is equal to x α = {x}, while x β may be written as x β = {x}. An observation of equations (4.17) and (4.18) learns that the space vector deals with a trans- formation process, in which a linear combination of the three supply variables x R ,x S ,x T , is converted to a two-phase x α ,x β form. It is important to realize that the space vector amplitude (|x|) and argument  arctan x β x α  can be a function of time. We may see non-continuous changes of both argument and amplitude in many cases such as three phase PWM. It is instructive at this stage to give an example based on equation (4.17) with C =1. In this case we will plot the space vector for three cases. Each [...]... generic module of this system according to figure 4. 7 It is at 92 FUNDAMENTALS OF ELECTRICAL DRIVES this stage helpful to recall the differential equation set of the circuit in question which is of the form diS1 dt diS2 = iS2 R + L dt diS3 = iS3 R + L dt uS1 = iS1 R + L (4. 41a) uS2 (4. 41b) uS3 (4. 41c) We can rewrite equations (4. 41) in a space vector form by making use of, for example, equation (4. 30) This... 0.0 04 0.006 0.008 0.01 time (s) 0.012 0.0 14 0.016 0.018 0.02 0 0.002 0.0 04 0.006 0.008 0.01 time (s) 0.012 0.0 14 0.016 0.018 0.02 40 0 uSβ (V) 200 0 −200 40 0 Figure 4. 33 Waveforms: uSα , uSβ 300 200 uSβ (V) 100 0 −100 −200 −300 40 0 −300 −200 Figure 4. 34 0 uSα (V) 100 200 300 Waveforms: uS123 , vector locus diagram m-file Tutorial 1, chapter 4 %tutorial 1, chapter 4 close all subplot(3,1,1) −100 40 0... transfer matrix is of the form given by equation (4. 49)    uDα uDβ = √  3 cos γ 4 sin γ 4 − sin γ 4 cos γ 4    uα uβ  (4. 49) 96 FUNDAMENTALS OF ELECTRICAL DRIVES The conversion module which converts the supply space vector format to phase variables uRST → uD1,D2,D3 is shown in figure 4. 22(b) Its contents can either be according to the set of generic modules shown in figure 4. 21(b) or the conversion... which upon substitution of equation (4. 16) may be written as uRST = C (uD1 − uD2 ) + (uD2 − uD3 ) ejγ + (uD3 − uD1 ) ej2γ 3 +CuS0 1 + ejγ + ej2γ (4. 43) 0 Equation (4. 42) may be rearranged by grouping the phase variables as shown in equation (4. 44)   uRST =     (4. 44)   C uD1 1−ej2γ +uD2 −1 + ejγ +uD3 −ejγ + ej2γ 3  √ jγ √ j ( γ +γ ) √ j ( γ +2γ ) 3e 4 3e 4 3e 4 The braced terms contain a common... by equations (4. 45) and (4. 56) The corresponding phasor relationships between the phase and supply based phasors is of the form uD123 = iD123 = √ γ 3 e−j 4 uRST γ 1 √ e−j 4 iRST 3 (4. 65a) (4. 65b) For the calculation of the current phasor iD123 use is made of equation (4. 63) (with subscript D123), in which uD123 is calculated using equation (4. 65a) Once the phasor iD123 is found equation (4. 65b) can be... transfer matrix of the form given by equation (4. 57)    iDα iDβ  1 3 = √  cos γ 4 sin γ 4 − sin γ 4 cos γ 4    iα  iβ (4. 57) The two conversion modules shown in figure 4. 23(b) can also be replaced by a single generic module as given in figure 4. 24( b) The corresponding transfer matrix as given by equation (4. 58) is found by combining the matrices represented by equations (4. 37), (4. 57) It is emphasized... plot(datoutn(:,6),datoutn(:,2)) grid xlabel(’time (s)’) ylabel(’u_{S\alpha} (V)’) axis([0 20e-3 -4 00 40 0]) subplot(2,1,2) plot(datoutn(:,6),datoutn(:,3),’r’) xlabel(’time (s)’) ylabel(’u_{S\beta} (V)’) axis([0 20e-3 -4 00 40 0]) grid figure plot(datoutn(:,2),datoutn(:,3)) axis equal axis( [ -4 00 40 0 -3 00 300]) grid xlabel(’u_{S\alpha} (V)’) ylabel(’u_{S\beta} (V)’) 4. 8.2 Tutorial 2 The relationship between supply, phase and space... as uS123 = dψS123 + iS123 R dt (4. 42) where ψS123 = LiS123 The development of the generic model proceeds along the lines discussed for the single phase R-L example A possible generic implementation of the three-phase system is given in figure 4. 19 The model Figure 4. 19 Generic, space vector based, model of three-phase R-L circuit (star connected) according to figure 4. 19 has as inputs the three phase... − xc ) (4. 35) xβ = C 2 Substitution of xc = − (xb + xa ) + 3x0 and use of (4. 34) gives after some manipulation xa = xb = − 1 1 xα + √ xβ + x0 3C C 3 (4. 36) The remaining variable xc is found by use of xc = − (xa + xb ) + 3x0 together with equations (4. 34) and (4. 35) The resultant complete conversion in matrix form as given by equation (4. 37), corresponds to the building block shown in figure 4. 17(b)... book refers to a so-called ‘power invariant’ notation form For this notational form the constant is chosen to be C = 2 The voltage and current space vectors given by equations (4. 22) 3 88 FUNDAMENTALS OF ELECTRICAL DRIVES and (4. 24) will then be of the form u = i = 3 u ejωt ˆ 2 3 ˆ j(ωt+ρ) ie 2 (4. 26a) (4. 26b) This means that the space vector amplitude of for example the voltage is of the form |u| = . i S3 =0 (4. 1a) u S1 + u S2 + u S3 =0 (4. 1b) Three-phase Circuits 77 Figure 4. 3. Three phase ma- chine, star (Y) connected Figure 4. 4. Star/Wye con- nected according figure 4. 3 Note that equation (4. 1b). observation of figure 4. 10 learns that the following two expressions may be found which contain the voltage u R . u R = u D1 + u T (4. 14a) u R = −u D2 + u S (4. 14b) Adding equations (4. 14a), (4. 14b) gives 2u R =. aid of equations (4. 1b), (4. 4) which leads to u S0 = u R + u S + u T 3 (4. 5) The conversion module whichrepresents equation (4. 4) is given byfigure 4. 6(a). An important observation from figure 4. 6(a)